Ask question

Ask question

All categories

3 years ago

12

Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

it reaches the bottom of the ramp

1 answer:

Arturiano [62]3 years ago

4 0

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s$

So, its velocity at the bottom of the ramp is 2.21 m/s.

You might be interested in

For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex

rjkz [21]

Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet

convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches

1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f

The focal length of the project lens is 3.84 inches

4 0

3 years ago

How do you calculate final velocity

serg [7]

Try looking google..

3 0

3 years ago

Read 2 more answers

Football player A has a mass of 210 pounds and is running at a rate of 5.0mi/hr. He collides with player B. Player B has a mass

LuckyWell [14K]

The answer is b.) the momentum before the collision is greater than the momentum after the collision

8 0

3 years ago

Read 2 more answers

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

Using calcium’s atomic structure, as shown in the image, what is this element’s atomic number?

ElenaW [278]

Answer:

Atomic number of calcium is 20.

6 0

3 years ago